Python - 将具有不同大小的列表的数组转换为字符串

Question

I have a Python array that contains a lot of lists, with different sizes.

myArray = [('Hello my name is ', ('Bond, James', 'Bond. It is ', '16:40', ' now'), '!!!')]

Question: What is the best way (more optimized) to get the bellow output, without creating a loop method which will append every value to a string?

I would like to see the following output:

"Hello my name is Bond, James Bond. It is 16:40 now!!!"

Answer 1

First, you have to flatten the nested list to a 'flat' list. Assuming that your list contains either strings or other lists (or tuples), you could use a function like this:

def flatten(lst):
    for x in lst:
        if isinstance(x, str):
            yield x
        else:
            for y in flatten(x):
                yield y

Incidentally, there also seems to be a function for this in the compiler.ast module, so alternatively you could just import that function and use it. However, this module is deprecated and has been removed in Python 3.

from compiler.ast import flatten

Either way, after the list has been flattened, you just have to join the segments to one string.

>>> list(flatten(myArray))
['Hello my name is ', 'Bond, James', 'Bond. It is ', '16:40', ' now', '!!!']
>>> ''.join(flatten(myArray))
'Hello my name is Bond, JamesBond. It is 16:40 now!!!'

Answer 2

You can write a very comprehensive recursive function like the one I wrote -

In [1]: def solve(x):
            if isinstance(x, str):
                return x
            return ''.join(solve(y) for y in x)

In [2]: solve(myArray)
Out[2]: 'Hello my name is Bond, JamesBond. It is 16:40 now!!!'

Answer 3

Here is a variation which joins as it flattens:

def deepJoin(stuff,d = ' '):
    if isinstance(stuff,str):
        return stuff
    else:
        return d.join(deepJoin(x,d) for x in stuff)

For example:

>>> deepJoin( [('Hello my name is ', ('Bond, James', 'Bond. It is ', '16:40', ' now'), '!!!')])
'Hello my name is  Bond, James Bond. It is  16:40  now !!!'

Answer 4

Another option is to use a list comprehension to flatten the list:

bond_says = ["".join(item) for sublist in myArray for item in sublist] # list comprehension
print(" ".join(bond_says))

Output:

"Hello my name is  Bond, JamesBond. It is 16:40 now !!!"

Answer 5

You could use regex (python 2.7):

import re
myArray = [('Hello my name is ', ('Bond, James', 'Bond. It is ', '16:40', ' now'), '!!!')]
print re.sub("(?:^|')[^']+(?:'|$)", '',str(myArray))

[Output]

Hello my name is Bond, JamesBond. It is 16:40 now!!!

UPDATE

Try this with myArray= [('Won\\'t work with this.')] – tobias_k 7 hours ago

import re
p = r"(?:^|(?<!\\)')[^'\"\\]+(?:'|$)|(?:^|\")[^\"]+(?:\"|$)"

myArray = [('Hello my name is ', ('Bond, James', 'Bond. It is ', '16:40', ' now'), '!!!')]
myArray2= [('Won\'t work with this.')]

print (re.sub(p, '', str(myArray)))
print (re.sub(p, '', str(myArray2)))

[Output]

Hello my name is Bond, JamesBond. It is 16:40 now!!!
Won't work with this.

Answer 6

A map makes it pretty concise:

def flatten(arr):
    if type(arr) == str:
        return arr.strip()
    else:
        return " ".join(map(flatten, arr))

returns "Hello my name is Bond, James Bond. It is 16:40 now !!!" for the example.